RRB Clerk Prelims Memory Based Paper (Held on:7 August 2022 Shift 1) (Previous Year Paper)

Statement:  A = B < C > D = E > G > I

Conclusion:

I. A > G → False (as per the statement A = B < C > D = E > G , There is no definite relation between A and G)

II. I < C → True ( as per the statement  C > D = E > G > I, C is greater than I)

Hence, Only II follows.

Statement: P < O ≤ U ≤ Y = R ≥ E ≥ B > A 

i) A < Y ⇒ True (as Y = R ≥ E ≥ B > A ⇒ Y > A or A < Y)

ii) R > P ⇒ True (as P < O ≤ U ≤ Y = R ⇒ P < R or R > P)

Hence, both I and II are true.

Given Statements: Q > B < R ≥ S < A = P 

Conclusions:

I. Q > S → False (as Q > B < R ≥ S < A = P  → No relation between them)

II. S > B → False (as Q > B < R ≥ S < A = P  → No relation between them)

Hence, neither conclusion I nor II are follows.

Given statements,

A > K ≤ O < J,  V ≥ H > A

On combining, we get:

V ≥ H > A > K ≤ O < J

Conclusions:

I. V > O → False (V ≥ H > A > K ≤ O → V > K ≤ O, a clear relation between V and O cannot be determined).

II. H > J → False (H > A > K ≤ O < J → H > K < J, a clear relation between H and J cannot be determined).

Hence, the correct answer is neither conclusion I nor II follows.

Given statements,

N < E ≤ K = P, B < F = K

On combining, we get:

N < E ≤ K = F > B, B < F = K = P

Conclusions:

I. F < N → False (N < E ≤ K = F → N < F i.e. F > N).

II. B < P → True (B < F = K = P → B < P).

Hence, the correct answer is only conclusion II follows.

1. P lives on prime numbered floor.

2. Three persons live between R and P’s floor.

3. N lives two floors above R’s floor.(so, Case 3 and case 4 are invalid)

4. O lives just above S’s floor.

5. Q lives above M’s floor.

6. M does not live on prime numbered floor. (So, case 2 would be invalid)

Hence, P lives on topmost floor.

1. P lives on prime numbered floor.

2. Three persons live between R and P’s floor.

3. N lives two floors above R’s floor.(so, Case 3 and case 4 are invalid)

4. O lives just above S’s floor.

5. Q lives above M’s floor.

6. M does not live on prime numbered floor. (So, case 2 would be invalid)

Hence, four persons live above R’s floor.

1. P lives on prime numbered floor.

2. Three persons live between R and P’s floor.

3. N lives two floors above R’s floor.(so, Case 3 and case 4 are invalid)

4. O lives just above S’s floor.

5. Q lives above M’s floor.

6. M does not live on prime numbered floor. (So, case 2 would be invalid)

Hence, only statement 3) is correct.

1. P lives on prime numbered floor.

2. Three persons live between R and P’s floor.

3. N lives two floors above R’s floor.(so, Case 3 and case 4 are invalid)

4. O lives just above S’s floor.

5. Q lives above M’s floor.

6. M does not live on prime numbered floor. (So, case 2 would be invalid)

Hence, Q lives three floors above R.

1. P lives on prime numbered floor.

2. Three persons live between R and P’s floor.

3. N lives two floors above R’s floor.(so, Case 3 and case 4 are invalid)

4. O lives just above S’s floor.

5. Q lives above M’s floor.

6. M does not live on prime numbered floor. (So, case 2 would be invalid)

Hence, number of persons living below R is equal to number of persons living above N.

Confusion Points

N lives two floors above R’s floor means there is a gap of one floor between N's floor and R's floor.

Given words are:

FAM MAN XTG TEM UUQ

Interchanging second and last letter of each word:

FMA, MNA, XGT, TME, UQU

Clearly, four words FMA, MNA, TME, UQU are ending with a vowel.

Hence, four words end with vowel.

Given words are:

FAM MAN XTG TEM UUQ

Arranging each letter in the given word in alphabetical order:

AFM, AMN, GTX, EMT, QUU

Clearly, GTX and QUU start with consonant.

Hence, two words will be started with a consonant.

Given words are:

FAM MAN XTG TEM UUQ

Adding “E” at the end of each word:

FAME, MANE, XTGE, TEME, UUQE

Clearly, FAME and MANE are two meaningful words.

Hence, two meaningful word will be formed.

Given words are:

FAM MAN XTG TEM UUQ

Changing all the consonant to immediate preceding letter and all the vowel to immediate succeeding letter as per English alphabetical series:

EBL, LBM, WSF, SFL, VVP

Clearly, only one word has a vowel i.e., EBL.

Hence, only one word will be formed.

Given words are:

FAM MAN XTG TEM UUQ

Arranging the given words in alphabetical order: FAM, MAN, TEM, UUQ, XTG

Clearly, two words FAM, MAN will be unchanged its position.

Hence, two words will remain unchanged.

Mistake Points 1. Here we have to count the words whose position not changed after the arrangement as given the questions.

Six Persons: H, I, J, K, L, and M.

1) Three people went on holiday between M and K.

2. M went after K.

3. Two people went on holiday between J and M.

4. H went on holiday immediately after L.

Only one position is left

5. I does not go in December.

Here, case 1 is invalid.

Hence, three persons went on holiday between H and I.

Six Persons: H, I, J, K, L, and M.

1) Three people went on holiday between M and K.

2. M went after K.

3. Two people went on holiday between J and M.

4. H went on holiday immediately after L.

Only one position is left

5. I does not go in December.

Here, case 1 is invalid.

Hence, K went on holiday on 5^th October.

Six Persons: H, I, J, K, L, and M.

1) Three people went on holiday between M and K.

2. M went after K.

3. Two people went on holiday between J and M.

4. H went on holiday immediately after L.

Only one position is left

5. I does not go in December.

Here, case 1 is invalid.

Hence, J went on a holiday on 2^nd November.

Six Persons: H, I, J, K, L, and M.

1) Three people went on holiday between M and K.

2. M went after K.

3. Two people went on holiday between J and M.

4. H went on holiday immediately after L.

Only one position is left

5. I does not go in December.

Here, case 1 is invalid.

Hence, three persons went on holiday before L.

Six Persons: H, I, J, K, L, and M.

1) Three people went on holiday between M and K.

2. M went after K.

3. Two people went on holiday between J and M.

4. H went on holiday immediately after L.

Only one position is left

5. I does not go in December.

Here, case 1 is invalid.

Hence, K went on holiday immediately after I.

The best possible Venn diagram is, 

Conclusions:

I. Some tablets are effective: False (No direct relation between tablets and effective, it is possible but not definite)

II. No effective is a tablets: False (No direct relation between effective and table, it is possible but not definite)

On complimentary conclusion I and II.

Hence, either conclusion I or II follows.

Key Points

Conditions for Either - Or 

1. Subject and predicate should be same
2. Both the individual conclusions should be false
3. One conclusion should be positive and one conclusion should be negative
4. There should either one of the following pairs present in the conclusions

• All + Some not
• Some + No

The least possible Venn diagram for the given statements is as follows

Conclusions: 

I. All A can be C → True  ( as there is no definite relation A and C is given, so its possibility is possible. hence, it is true) 

II. Some C is A  →  False  ( as there is no relation between A and C is given, it is possible but not definite. hence it is false)

Hence, Only I follow.

The least possible diagram is given below

Conclusion:

I. Some Salty is Pure → False (There is no definite relation between Salty and Pure but it is possible. So, it is false)

II. No Pure is Salty → False (There is no definite relation between Salty and Pure but it is possible. So, it is false)

Hence, Either I or II follows.

Points to remember:

There are only two complementary pairs for either - or case:

A. Some - No

B. All - Some not

Conditions for either - or:-

1. Both the conclusions must be individually false but not definitely false.
2. Conclusion should be Complementary Pair.
3. In the case of Some - No subject and predicate can interchange their positions.

The least possible Venn diagram for the given statements is as follows:

Conclusions:

I. Some Ball being Red is a possibility → True (As there is no direct relation between Ball and Red some ball being red is possible but not definite. So, it is true)

II. Some Ball is Red → False (As there is no direct relation between Ball and Red it is possible but not definite. So, it is false)

Hence, Only I follow.

The least possible Venn diagram for the given statements is as follows:

Conclusion:

I. No Walking is Jogging → False (There is no definite relation between Walking and Jogging. So, it is false)

II. Some Walking is Jogging → False (There is no definite relation between Walking and Jogging. So, it is false)

Hence, Either I or II follows.

Points to remember:

There are only two complementary pairs for either - or case:

A. Some - No

B. All - Some not

Conditions for either - or:-

1. Both the conclusions must be individually false but not definitely false.
2. Conclusion should be Complementary Pair.
3. In the case of Some - No subject and predicate can interchange their positions.

Persons: A, B, C, D, E, F and G

1) A is neighbor of B and C.

2) D is not neighbor of C or E.

3) G, who is second to the left of D, is the neighbor of E and F.

In case 1 the above condition is not satisfied in this case and thus eliminated.

The above arrangement will be final arrangement.

Hence, D sits second to left of A.

Persons: A, B, C, D, E, F and G

1) A is neighbor of B and C.

2) D is not neighbor of C or E.

3) G, who is second to the left of D, is the neighbor of E and F.

In case 1 the above condition is not satisfied in this case and thus eliminated.

The above arrangement will be final arrangement.

Thus, E is second to the left of F.

Persons: A, B, C, D, E, F and G

1) A is neighbor of B and C.

2) D is not neighbor of C or E.

3) G, who is second to the left of D, is the neighbor of E and F.

In case 1 the above condition is not satisfied in this case and thus eliminated.

The above arrangement will be final arrangement.

E is the neighbour of G and C.

Thus, "E is the neighbour of G and F" is the incorrect statement.

Persons: A, B, C, D, E, F and G

1) A is neighbor of B and C.

2) D is not neighbor of C or E.

3) G, who is second to the left of D, is the neighbor of E and F.

In case 1 the above condition is not satisfied in this case and thus eliminated.

The above arrangement will be final arrangement.

Thus, G sits third to right of A.

Persons: A, B, C, D, E, F and G

1) A is neighbor of B and C.

2) D is not neighbor of C or E.

3) G, who is second to the left of D, is the neighbor of E and F.

In case 1 the above condition is not satisfied in this case and thus eliminated.

The above arrangement will be final arrangement.

Thus, F is second to left of B.

Given Word: SOURCING

The given word can be represented as follows:

Thus, there are only 2 such pairs → Forward: SU; Backward: GI.

Hence, 2 is the correct answer.

The first, third, sixth and eighth letters of the word ‘ACCOUTERED are A, C, T and R.

A meaningful word from this is - CART

Hence, the correct answer is R.

Given series: 

LEFT END  L 4 B Δ W U S = 8 C N F 5 X T @ K O 1 Q V D β 6 A P E J R H + M Z  RIGHT END

Numbers that are immediately preceded by a symbol and immediately followed by a vowel:

LEFT END  L 4 B Δ W U S = 8 C N F 5 X T @ K O 1 Q V D β 6 A P E J R H + M Z  RIGHT END

Clearly, only 6 is immediately preceded by a symbol and also immediately followed by a vowel.

Hence, 1 is the correct answer.

Given Series: L 4 B Δ W U S = 8 C N F 5 X T @ K O 1 Q V D β 6 A P E J R H + M Z

In each group, the pattern follows as:

The second element is second to the left of the first element.

The third element is third to the right of the second element.

Also in the series,

the first element of the second group (i.e. U of UΔS) is the immediate right of the third element of the first group (i.e. W of Δ4W)

LEFT END  L 4 B Δ W U S = 8 C N F 5 X T @ K O 1 Q V D β 6 A P E J R H + M Z  RIGHT END

By following the above pattern we get FC5.

Hence, FC5 is the correct answer.

Given series: 

LEFT END  L 4 B Δ W U S = 8 C N F 5 X T @ K O 1 Q V D β 6 A P E J R H + M Z   RIGHT END

Clearly, there is no such consonant that is immediately preceded by a symbol and followed by a number.

Hence, 0 is the correct answer.

Given series: 

LEFT END  L 4 B Δ W U S = 8 C N F 5 X T @ K O 1 Q V D β 6 A P E J R H + M Z  RIGHT END

Product of all the numbers:

LEFT END  L 4 B Δ W U S = 8 C N F 5 X T @ K O 1 Q V D β 6 A P E J R H + M Z  RIGHT END 

4×8×5×1×6 = 960

Total number of vowels in the series:

LEFT END  L 4 B Δ W U S = 8 C N F 5 X T @ K O 1 Q V D β 6 A P E J R H + M Z  RIGHT END 

There are 4 vowels: U, O, A, E

So, 960 ÷ 4 = 240

Hence, 240 is the correct answer.

Given series: 

LEFT END  L 4 B Δ W U S = 8 C N F 5 X T @ K O 1 Q V D β 6 A P E J R H + M Z  RIGHT END

By placing 7 elements from the right end between the 8^th and 9^th element from the left end.

LEFT END  L 4 B Δ W U S = E J R H + M Z 8 C N F 5 X T @ K O 1 Q V D β 6 A P  RIGHT END

7^th element in the new arrangement from the left end:

LEFT END  L 4 B Δ W U S = E J R H + M Z 8 C N F 5 X T @ K O 1 Q V D β 6 A P  RIGHT END

Clearly, S is the seventh element from the left end after the arrangement.

Hence, S is the correct answer.

Below table represents symbols used to draw a family tree,

1. S is mother of O, who is mother of Q.

2. N is mother of M.

3. S is married.

4. Q’s father is son-in-law of M. So, M must be husband of S.

5. The gender of P and Q is same. So, P must be father of Q, who must be male member.

We can draw the following Family Tree from the information given above:

Hence, N is Mother-in-law of S.

Below table represents symbols used to draw a family tree,

1. S is mother of O, who is mother of Q.

2. N is mother of M.

3. S is married.

4. Q’s father is son-in-law of M. So, M must be husband of S.

5. The gender of P and Q is same. So, P must be father of Q, who must be male member.

We can draw the following Family Tree from the information given above:

Hence, O is daughter of S.

Below table represents symbols used to draw a family tree,

1. S is mother of O, who is mother of Q.

2. N is mother of M.

3. S is married.

4. Q’s father is son-in-law of M. So, M must be husband of S.

5. The gender of P and Q is same. So, P must be father of Q, who must be male member.

We can draw the following Family Tree from the information given above:

  Hence, M is grandfather of Q.

Given:

The total number of Almonds sold = 50

The total number of Raisins sold = 60

The total number of Cashews sold = 80

The total number of Peanuts sold = 110

The total number of Walnuts sold = 90

Calculation:

The required ratio = (Almonds + Cashews) : Peanuts

⇒ (50 + 80) : 110

⇒ 130 : 110

⇒ 13 : 11

∴ The required ratio is 13 : 11.

Given:

The total number of Almonds sold = 50

The total number of Raisins sold = 60

The total number of Cashews sold = 80

The total number of Peanuts sold = 110

The total number of Walnuts sold = 90

Calculation:

The total number of dry fruits sold = Almonds + Raisins + Cashews + Peanuts + Walnuts

⇒ 50 + 60 + 80 + 110 + 90 = 390

∴ The total number of dry fruits sold by Walmart in a particular month is 390.

Given:

The total number of Almonds sold = 50

The total number of Raisins sold = 60

The total number of Cashews sold = 80

The total number of Peanuts sold = 110

The total number of Walnuts sold = 90

Calculation:

The required difference = (Peanuts  + Walnuts) - (Raisins + Cashews)

⇒ (110 + 90) - (60 + 80)

⇒ 200 - 140

⇒ 60

∴ The required difference is 60.

Given:

The total number of Almonds sold = 50

The total number of Raisins sold = 60

The total number of Cashews sold = 80

The total number of Peanuts sold = 110

The total number of Walnuts sold = 90

Formula Used:

Average = Sum of observations/Total number of observations

Calculation:

The total number of dry fruits sold =  50 + 60 + 80 + 110 + 90

⇒ 390

The required average = 390/5

⇒ 78

∴ The required average is 78.

Given:

The total number of Almonds sold = 50

The total number of Raisins sold = 60

The total number of Cashews sold = 80

The total number of Peanuts sold = 110

The total number of Walnuts sold = 90

Calculation:

The required % = [(90 - 50)/50] × 100

⇒ [40/50] × 100 = 80%

∴ The required answer is 80%

Given:

Average of 5 numbers = 28

Average of first three numbers = 18

Concept used:

Average = (sum of all terms)/total number of terms

Sum of all terms = Average × total number of terms

Calculation:

Sum of 5 numbers = 5 × 28 = 140

Sum of first 3 numbers = 3 × 18 = 54

Sum of remaining numbers = 140 – 54 = 86

Number of remaining numbers = 2

Average of remaining numbers = 86/2 = 43

∴ The average of remaining numbers is 43

Shortcut Trick

Radha bought an article at C.P. If she gave D% discount she still manages to earn P% of the profit.

The marked price of the article = C.P × \(\frac{100\text {%} + P\text {%}}{100\text {%} - D\text {%}}\)

According to the question,

C.P = ₹100

P = 12.5%

D = 25%

Now,

M.P = ₹100 × \(\frac{100\text {%} + 12.5\text {%}}{100\text {%} - 25\text {%}}\) = ₹150

∴ The marked price of the article is ₹150.

Alternate Method

Given:

Radha bought an article at ₹100.

If she gave 25% discount she still manages to earn 12.5% of profit.

Formula used:

(1) Selling Price = (100% - D%) of Marked Price

Where, 

D, is the value of the discount per cent

(2) Cost price = \(\frac{\text {Selling Price}}{100\text {%} + \text {Profit %}}\)

Calculation:

The cost price of the article = ₹100

The profit earned = 12.5%

Therefore,

The selling price of the article = (100% + 12.5%) of ₹100 = ₹112.5

Let 'M' is the marked price of the article.

The discount offered = 25%

Therefore, 

⇒ 112.5 = (100% - 25%) of MP

⇒ MP = ₹150

∴ The marked price of the article is ₹150.

Given:

The ratio of the present ages of A and B = 7 : 11

After 10 years the ratio become = 13 : 19

Calculation:

Let the present ratio of A and B be 7x and 11x respectively.

A/B = (7x + 10)/(11x + 10) = 13/19

⇒ 133x + 190 = 143x + 130

⇒ 10x = 60

⇒ x = 6

Present age of B = 11x

⇒ 11 × 6

⇒ 66

∴ The present age of B is 66 years.

Alternate Method

The difference between the ages of A and B will always be the same, so we equalize the difference between the two ratios. 

The ratio of the present ages of A and B = 7 : 11

After 10 years the ratio become = 13 : 19

For equalizing the difference between the two ratios, we can multiple with 3 in present ratio and multiple with 2 in after 10 years ratio:

⇒ 21 : 33

⇒ 26 : 38

⇒ 26 - 21 or 38 - 33 = 10 Years

⇒ 5 unit = 10 years

⇒ 1 unit = 2 years

The present age of B = 33 unit

⇒ 33 × 2

⇒ 66

∴ The present age of B is 66 years.

Given:

44 × 25% of 1331 ÷ 11 = ?

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

44 × 25% of 1331 ÷ 11 = ?

⇒ 44 × (1/4) of 1331 ÷ 11 = ?

⇒ 44 × (1331/4) × (1/11) = ?

⇒ ? = 1331

∴ The value of ? will be 1331.

Given:

(20 + 5)% of 360 = ? - 10% of 102 

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

(20 + 5)% of 360 = ? - 10% of 102 

⇒ 25% of 360 = ? - 10% of 100

⇒ 90 = ? - 10

⇒ ? = 100

∴ The value of ? is 100.

Calculation:

 7\(\frac{1}{2}\)- 6\(\frac{3}{4}\) + 5\(\frac{1}{4}\) = ?

⇒ \(\frac{15}{2}-\frac{27}{4}+\frac{21}{4}\) = ?

⇒ \(\frac{15\times 2-27+21}{4}\) = ?

⇒ \(\frac{30-6}{4}=\frac{24}{4}\) 

⇒ 6

∴ The required value is 6.

Additional Information BODMAS Rule:

Given:

√676 + √576 = (5)2 × ?

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

√676 + √576 = (5)2 × ?

⇒ 26 + 24 = 25 × ?

⇒ 50 = 25 × ?

⇒ ? = 2

∴ The value of ? is 2.

Given:

(√225 × 12) ÷ 45 + 19 × 5 = ?

Concept Used:

Follow the BODMAS rule according to the table given below:

Calculation:

(√225 × 12) ÷ 45 + 19 × 5 = ?

⇒ (15 × 12) ÷ 45 + 95 = ?

⇒ 180 ÷ 45 + 95 = ?

⇒ 4 + 95 = ?

⇒ ? = 99

∴ The value of ‘?’ Is 99.

Hint

√225 = 15

225 = (15 × 15) or (15)2

Given:

95 – 10 × [15 × (80 – 75)] ÷ 25 = ?

Concept Used:

Follow the BODMAS rule according to the table given below:

Calculation:

95 – 10 × [15 × (80 – 75)] ÷ 25 = ?

⇒ 95 – 10 × (15 × 5) ÷ 25 = ?

⇒ 95 - 10 × (75/25) = ?

⇒ 95 - 30 = ?

⇒ ? = 65

∴ The value of ‘?’ is 65.

Given:

8909.87 ÷ 300.14 × 10.23 = ?

Concept used:

Calculation:

8909.87 ÷ 300.14 × 10.23 = ?

⇒ 8910 ÷ 300 × 10 = ?

⇒ ? = 297 

∴ The value of ? is 297.

Given:

15 × 252 ÷ 60 + 170 = ? + 63

Concept Used:

Calculation:

15 × 252 ÷ 60 + 170 = ? + 63

⇒ 63 + 170 = ? + 63

⇒ ? = 170

∴ The value of ? is 170.

Given:

√(62 + 82) + 24 = 2 × √?

Concept Used:

Follow the BODMAS rule according to the table given below:

Calculation:

√(62 + 82) + 24 = 2 × √?

⇒ √(36 + 64) + 24 = 2 × √?

⇒ √100 + 24 = 2 × √?

⇒ 10 + 24 = 2 × √?

⇒ 34/2 = √?

⇒ 17 = √?

⇒ ? = (17)²

⇒ ? = 289

∴ 289 will come in place of the question mark ('?').

Given:  

20% of 200 + 40% of 400 = ?

Concept Used:

Follow the BODMAS rule according to the table given below:

Calculation:

20% of 200 + 40% of 400 = ?

⇒ (20/100) × 200 + (40/100) × 400 = ?

⇒ 20 × 2 + 40 × 4 = ?

⇒ 40 + 160 = ?

⇒ ? = 200

∴ 200 will come in place of the question mark ('?').

Given:

32% of 200 × 53% of ? = 6784

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

32% of 200 × 53% of ? = 6784

⇒ 64 × 53% of ? = 6784

⇒ 53% of ? = 6784/64

⇒ 53% of ? = 106

⇒ ? = (106/53) × 100

⇒ ? = 200

∴ The value of ? is 200.

Given:

62.5% of 128 + {(√36)/3} × 4 = ? + 10

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

62.5% of 128 + {(√36)/3} × 4 = ? + 10

⇒ (5/8) × 128 + {6/3} × 4 = ? + 10

⇒ 80 + 8 = ? + 10

⇒ 88 = ? + 10

⇒ ? = 78

∴ The value of ? is 78.

Hint

Given:  

25% of 840 + √49 × 7 = (?)4 + ∛27

Concept Used:

Follow the BODMAS rule according to the table given below:

Calculation:

25% of 840 + √49 × 7 = (?)4 + ∛27

⇒ (25/100) × 840 + 7 × 7 = (?)⁴ + 3

⇒ (840/4) + 49 = (?)⁴ + 3

⇒ 210 + 49 - 3 = (?)⁴

⇒ 256 = (?)⁴

⇒ (4)⁴ = (?)⁴

⇒ ? = 4

∴ 4 will come in place of the question mark ('?').

Important Points

If (x)^a = (y)^a then x = y.

Given:

17 × √9 + 30% of 200 = 25% of 80 + ? × 13

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

17 × √9 + 30% of 200 = 25% of 80 + ? × 13

⇒ 17 × 3 + (30/100) × 200 = (25/100) × 80 + ? × 13

⇒ 51 + 60 = 20 + ? × 13

⇒ 111 – 20 = ? × 13

⇒ 91 = ? × 13

⇒ ? = 7

∴ The value of ? is 7

Given:

11² + 120 ÷ 12 × 2 = ? – 30

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

11² + 120 ÷ 12 × 2 = ? – 30

⇒ 121 + 10 × 2 = ? - 30 

⇒ 121 + 20 = ? - 30 

⇒ ? = 141 + 30 = 171

⇒ ? = 171

∴ The value of ? is 171.

Given:

Rate = 5%

Time = 2 years

Difference of SI and CI = 400

Formula Used:

Sum = Difference × 100²/r²

Calculation:

To solve this question, we can apply a short trick approach

Sum = Difference × 100²/r²

Difference = ₹ 400, r = 5%

By the short trick approach, we get

∴ Sum = 400 × 100²/5²             

⇒ 1,60,000

∴ The sum is Rs. 1,60,000.

We can solve it by the net% formula,

Rate % of SI for 2 years at 5% pa = 5 × 2

⇒ 10%

∴ Rate % of CI for 2 years at 5%,

⇒ 5 + 5 + (5 × 5)/100

⇒ 10.25%

% rate difference of CI and SI = 10.25% – 10% 

⇒ 0.25%

Let the sum be ₹ x, then

0.25% of x = 400

x = 400 × 100/0.25      

⇒ 1,60,000

∴ The sum is Rs. 1,60,000.

Given:

A train running at 90 km/h crosses an electric pole in 18 seconds and a platform in 65 seconds.

Concept used:

1m/s = 18/5 km/h

While a train crosses a platform, it crosses the sum of its length and the length of the platform

While a train crosses an electric pole, it crosses its own length.

Calculation:

Let the length of the platform be P meters.

90 kmph = 90 ÷ 18/5 = 25 meters per second (mps)

Hence, the length of the train = 25 × 18 = 450 meters.

According to the question,

25 × 65 = (450 + Q)

⇒ 1625 = 450 + Q

⇒ Q = 1175

∴ The length of the platform is 1175 m.

Let ‘l’, ’b’ be the length and breadth of the rectangle.

 Length of the rectangle = l = 7x

Breadth of the rectangle = b = 3x

Breadth is 28 metre less than the Length

⇒ b = l – 28

⇒ 3x = 7x - 28

⇒ 28 = 7x - 3x

⇒ 4x = 28

⇒ x = 28/4 = 7 metres

∴ Length of the rectangle = l = 7x = 7 × 7 = 49 metres

∴ Breadth of the rectangle = b = 3x = 3 × 7= 21 metres

Perimeter of the rectangle = 2(l + b)

⇒ Perimeter of the rectangle = 2 × (49 + 21)

⇒ Perimeter of the rectangle = 2 × 70 = 140 metres

∴ Perimeter of the rectangle = 140 metres

Given:

Aman ’s share : Ravi’s share = (Aman ’s investment : Ravi’s investment) = 24000 : 8000 = 3 : 1

Calculation:

Let be Aman ’s share = 3x and Ravi 's share = 1x

According to the question  

3x + x = 48000

⇒ 4x = 48000

⇒ x = 12000

Put the value of x in Aman  & Ravi 

Share of B = Rs. 12000 

Share of Aman = 3x = 3 × 12000 = Rs. 36000.

∴ Share of A  and Ravi  is  Rs. 36,000 &  Rs. 12,000.